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Evaluate the integral of ((x-a)(b-x))-1/2

Evaluate the following integral for b \neq a.

    \[ \int \frac{dx}{\sqrt{(x-a)(b-x)}}. \]

(Note: Again, there is an error in the answers in the back of the book for this problem and the previous one. The answers listed for problems #46 and #47 should be swapped.)

Following the hint, we make the substitution

    \begin{align*}  && x-a &= (b-a) \sin^2 u \\ \implies && b-x &= (b-a) (1- \sin^2 u) \\ \implies && u &= \arcsin \sqrt{\frac{x-a}{b-a}} \\ \implies && dx &= 2(b-a) \sin u \cos u \, du. \end{align*}

Therefore, we can rewrite this integral in terms of u,

    \begin{align*}  \int \frac{dx}{\sqrt{(x-a)(b-x)}} &= \int \frac{2(b-a) \sin u \cos u \, du}{\sqrt{(b-a)^2 \sin^2 u (1 - \sin^2 u)}} \\  &= \frac{2(b-a)}{|b-a|} \int \frac{\cos u \, du}{\sqrt{1-\sin^2 u}} \\  &= \frac{2(b-a)}{|b-a|} \int du \\  &= \frac{2(b-a)}{|b-a} u + C. \end{align*}

Substituting back in for u we then have

    \begin{align*}  \int \frac{dx}{\sqrt{(x-a)(b-x)}} &= \frac{2(b-a)}{|b-a|} u + C \\  &= \frac{2(b-a)}{|b-a|} \arcsin \sqrt{\frac{x-a}{b-a}} + C. \end{align*}

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